Find the square root of
(a) `-8-6i`
(b) `(x^2)/(y^2)+ (y^2)/(x^2)-(1)/(i) ((x)/(y)-(y)/(x))-(9)/(4)`.

Answer

To find the square root of the given complex numbers, we will solve each part step by step.

### Part (a): Find the square root of `-8 - 6i`

1. **Express the complex number in the form of \( a + bi \)**:
We have \( z = -8 - 6i \).

2. **Assume the square root is in the form \( x + yi \)**:
Let \( w = x + yi \), where \( x \) and \( y \) are real numbers.

3. **Square the assumed square root**:
\[ w^2 = (x + yi)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + (2xy)i \]

4. **Set the real and imaginary parts equal to those of \( z \)**:
\[ x^2 - y^2 = -8 \quad \text{(1)} \]
\[ 2xy = -6 \quad \text{(2)} \]

5. **From equation (2), express \( y \) in terms of \( x \)**:
\[ y = \frac{-6}{2x} = \frac{-3}{x} \quad \text{(3)} \]

6. **Substitute (3) into (1)**:
\[ x^2 - \left(\frac{-3}{x}\right)^2 = -8 \]
\[ x^2 - \frac{9}{x^2} = -8 \]

7. **Multiply through by \( x^2 \) to eliminate the fraction**:
\[ x^4 + 8x^2 - 9 = 0 \]

8. **Let \( u = x^2 \)**:
\[ u^2 + 8u - 9 = 0 \]

9. **Use the quadratic formula to solve for \( u \)**:
\[ u = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm 10}{2} \]
\[ u = 1 \quad \text{or} \quad u = -9 \quad \text{(discard negative)} \]

10. **Find \( x \)**:
\[ x^2 = 1 \implies x = 1 \quad \text{or} \quad x = -1 \]

11. **Find \( y \) using (3)**:
- If \( x = 1 \):
\[ y = \frac{-3}{1} = -3 \]
- If \( x = -1 \):
\[ y = \frac{-3}{-1} = 3 \]

12. **Thus, the square roots are**:
\[ \sqrt{-8 - 6i} = 1 - 3i \quad \text{or} \quad -1 + 3i \]

### Part (b): Find the square root of \( \frac{x^2}{y^2} + \frac{y^2}{x^2} - \frac{1}{i} \left( \frac{x}{y} - \frac{y}{x} \right) - \frac{9}{4} \)

1. **Simplify the expression**:
\[ \frac{x^2}{y^2} + \frac{y^2}{x^2} - \frac{1}{i} \left( \frac{x}{y} - \frac{y}{x} \right) - \frac{9}{4} \]

2. **Rewrite \( \frac{1}{i} \)**:
\[ \frac{1}{i} = -i \quad \text{(since \( i \cdot -i = 1 \))} \]
Thus,
\[ -\frac{1}{i} \left( \frac{x}{y} - \frac{y}{x} \right) = i \left( \frac{x}{y} - \frac{y}{x} \right) \]

3. **Combine terms**:
\[ \frac{x^2}{y^2} + \frac{y^2}{x^2} + i \left( \frac{x}{y} - \frac{y}{x} \right) - \frac{9}{4} \]

4. **Let \( a = \frac{x^2}{y^2} + \frac{y^2}{x^2} - \frac{9}{4} \) and \( b = \frac{x}{y} - \frac{y}{x} \)**:
\[ z = a + ib \]

5. **Find the square root**:
Assume \( w = u + iv \):
\[ w^2 = (u + iv)^2 = u^2 - v^2 + 2uvi \]

6. **Set real and imaginary parts equal**:
\[ u^2 - v^2 = a \quad \text{(4)} \]
\[ 2uv = b \quad \text{(5)} \]

7. **From (5), express \( v \) in terms of \( u \)**:
\[ v = \frac{b}{2u} \quad \text{(6)} \]

8. **Substitute (6) into (4)**:
\[ u^2 - \left(\frac{b}{2u}\right)^2 = a \]
\[ u^2 - \frac{b^2}{4u^2} = a \]

9. **Multiply through by \( 4u^2 \)**:
\[ 4u^4 - b^2 = 4au^2 \]
\[ 4u^4 - 4au^2 - b^2 = 0 \]

10. **Let \( v = u^2 \)**:
\[ 4v^2 - 4av - b^2 = 0 \]

11. **Use the quadratic formula**:
\[ v = \frac{4a \pm \sqrt{(4a)^2 + 16b^2}}{8} = \frac{a \pm \sqrt{a^2 + 4b^2}}{2} \]

12. **Find \( u \) and \( v \)**:
\[ u = \sqrt{v} \quad \text{and} \quad v = \frac{b}{2u} \]

13. **Thus, the square root is**:
\[ \sqrt{\frac{x^2}{y^2} + \frac{y^2}{x^2} - \frac{1}{i} \left( \frac{x}{y} - \frac{y}{x} \right) - \frac{9}{4}} = \pm \left( u + iv \right) \]